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alamsal/distanceFinder.js

Created August 31, 2015 18:55
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Find out location with in X radius from the user's current location
Raw
distanceFinder.js
function distance(obj) {
var R = 6371; // km
var dLat = (obj.lat2 - obj.lat1) * Math.PI / 180;
var dLon = (obj.lon2 - obj.lon1) * Math.PI / 180;
var lat1 = obj.lat1 * Math.PI / 180;
var lat2 = obj.lat2 * Math.PI / 180;
var a = Math.sin(dLat/2) * Math.sin(dLat/2) +
Math.sin(dLon/2) * Math.sin(dLon/2) * Math.cos(lat1) * Math.cos(lat2);
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
var d = R * c;
var m = d * 0.621371;
return {
km: d,
m: m
}
}
var location = {
lat: 51.5171,
lon: -0.1062,
title: 'London'
};
var points = [
{ lat: 52.4800, lon: -1.9100, title: 'Birmingham' },
{ lat: 53.7500, lon: -0.3600, title: 'Hull' },
{ lat: 52.2100, lon: 0.1300, title: 'Cambridge' },
{ lat: 51.9000, lon: -0.2000, title: 'Stevenage' },
{ lat: 51.1092, lon: -0.1872, title: 'Crawley' }
];
var filtered = [], intPoint = 0, intPoints = points.length, dist = {}, point = {};
for (intPoint = 0; intPoint < intPoints; intPoint = intPoint + 1) {
point = points[intPoint];
dist = distance({
lat1: location.lat,
lon1: location.lon,
lat2: point.lat,
lon2: point.lon
});
if (dist.km < 100) {
point.distance = dist.km;
filtered.push(point);
}
}
Ti.API.info('points: ' + JSON.stringify(filtered));
//https://developer.appcelerator.com/question/149231/find-out-location-in-10-km-radius-from-user-current-location
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